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Consider two concentric toruses, and let $\Sigma$ be the domain interior to the greater torus and exterior to the smaller torus. Is it possible to find a vector field $\mathbf{b}$ satisfying the following conditions:

  1. $\mathrm{div} \, \mathbf{b}=0$ in $\Sigma$;

  2. $\mathbf{b}$ is everywhere tangent to $\partial \Sigma$; and

  3. $\mathrm{curl} \, \mathbf{b} = \lambda(\mathbf{x}) \mathbf{b}$ in $\Sigma$.

Thank you

Muphrid
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user48900
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  • Is $\lambda$ prescribed to us? If yes, what do we know about it? If no, have you tried to find a field with zero curl and divergence? –  Jan 02 '13 at 23:59
  • No $\lambda$ is an arbitrary continuous function. In Wikipedia (page on Hairy ball theorem) there is a picture of field everywhere tangent to the torus, but I don't know if we can extend it to be Beltrami. – user48900 Jan 03 '13 at 00:02
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    Would not the two-dimensional example $\dfrac{x_2\mathbf{i}-x_1\mathbf{j}}{x_1^2+x_2^2}$ work just as well in three dimensions? –  Jan 03 '13 at 00:10
  • Maybe after you compose it with a map $\pi$ which sends $\mathbb{R}^2$ to the torus. – user48900 Jan 03 '13 at 00:20
  • No, I mean $F(x)=\dfrac{x_2\mathbf{i}-x_1\mathbf{j}}{x_1^2+x_2^2}$ for all $x\in \mathbb R^3$. Does not this satisfy 1,2,3? I'm taking the $x_3$-axis to be the axis of symmetry of $\Sigma$. –  Jan 03 '13 at 00:31
  • No. How can this two dimensional field be everywhere tangent to the torus, which does not lie in a single plane? – user48900 Jan 03 '13 at 00:33
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    I don't why not. The flow lines of $F$ are circles centered on the $x_3$-axis. The domain is rotationally symmetric. Hence, flow lines do not exit the domain. –  Jan 03 '13 at 00:35
  • Yes, you are right. Is this Beltrami? – user48900 Jan 03 '13 at 00:38
  • The curl is zero, and so is the divergence... –  Jan 03 '13 at 00:39
  • And in the case of $\lambda$ prescribed, what do you do? – user48900 Jan 03 '13 at 00:44
  • Delete my comments and abandon this thread, I suppose. –  Jan 03 '13 at 00:47

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